What is the solution to the given inequality?<br><br> -2x + 4

How to solve 2(x+y)^2-9(x+y)-5

Aliun [14]

Answer:

2(x + y)² - 9( x + y ) -5 = 0

⇒2(x + y)² - 10 (x+y) +1(x+y) -5 = 0

⇒2(x+y)(x + y - 5 ) + 1(x + y -5 ) = 0

taking (x + y -5 ) common ,

⇒(x + y -5 )[2(x + y) + 1] =0

⇒(x + y -5)(2x + 2y +1) =0

hope , you got this

6 0

4 years ago

Solve for x<br> 6/x^2+2x-15 +7/x+5 =2/x-3

timama [110]

Answer:

x = ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3))/(15 (2140 - 9 sqrt(56235))^(1/3)) - 1/3 or x = 1/15 (17 5^(2/3) (-1/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) - 1/3 or x = -1/3 - 17/(3 (10700 - 45 sqrt(56235))^(1/3)) - (2140 - 9 sqrt(56235))^(1/3)/(3 5^(2/3))

Step-by-step explanation:

Solve for x:

6/x^2 + (2 x - 8)/(x + 5) = 2/x - 3

Bring 6/x^2 + (2 x - 8)/(x + 5) together using the common denominator x^2 (x + 5). Bring 2/x - 3 together using the common denominator x:

(2 (x^3 - 4 x^2 + 3 x + 15))/(x^2 (x + 5)) = (2 - 3 x)/x

Cross multiply:

2 x (x^3 - 4 x^2 + 3 x + 15) = x^2 (2 - 3 x) (x + 5)

Expand out terms of the left hand side:

2 x^4 - 8 x^3 + 6 x^2 + 30 x = x^2 (2 - 3 x) (x + 5)

Expand out terms of the right hand side:

2 x^4 - 8 x^3 + 6 x^2 + 30 x = -3 x^4 - 13 x^3 + 10 x^2

Subtract -3 x^4 - 13 x^3 + 10 x^2 from both sides:

5 x^4 + 5 x^3 - 4 x^2 + 30 x = 0

Factor x from the left hand side:

x (5 x^3 + 5 x^2 - 4 x + 30) = 0

Split into two equations:

x = 0 or 5 x^3 + 5 x^2 - 4 x + 30 = 0

Eliminate the quadratic term by substituting y = x + 1/3:

x = 0 or 30 - 4 (y - 1/3) + 5 (y - 1/3)^2 + 5 (y - 1/3)^3 = 0

Expand out terms of the left hand side:

x = 0 or 5 y^3 - (17 y)/3 + 856/27 = 0

Divide both sides by 5:

x = 0 or y^3 - (17 y)/15 + 856/135 = 0

Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later:

x = 0 or 856/135 - 17/15 (z + λ/z) + (z + λ/z)^3 = 0

Multiply both sides by z^3 and collect in terms of z:

x = 0 or z^6 + z^4 (3 λ - 17/15) + (856 z^3)/135 + z^2 (3 λ^2 - (17 λ)/15) + λ^3 = 0

Substitute λ = 17/45 and then u = z^3, yielding a quadratic equation in the variable u:

x = 0 or u^2 + (856 u)/135 + 4913/91125 = 0

Find the positive solution to the quadratic equation:

x = 0 or u = 1/675 (9 sqrt(56235) - 2140)

Substitute back for u = z^3:

x = 0 or z^3 = 1/675 (9 sqrt(56235) - 2140)

Taking cube roots gives (9 sqrt(56235) - 2140)^(1/3)/(3 5^(2/3)) times the third roots of unity:

x = 0 or z = (9 sqrt(56235) - 2140)^(1/3)/(3 5^(2/3)) or z = -((-1)^(1/3) (9 sqrt(56235) - 2140)^(1/3))/(3 5^(2/3)) or z = ((-1)^(2/3) (9 sqrt(56235) - 2140)^(1/3))/(3 5^(2/3))

Substitute each value of z into y = z + 17/(45 z):

x = 0 or y = (9 sqrt(56235) - 2140)^(1/3)/(3 5^(2/3)) - (17 (-1)^(2/3))/(3 (5 (2140 - 9 sqrt(56235)))^(1/3)) or y = 17/3 ((-1)/(5 (2140 - 9 sqrt(56235))))^(1/3) - ((-1)^(1/3) (9 sqrt(56235) - 2140)^(1/3))/(3 5^(2/3)) or y = ((-1)^(2/3) (9 sqrt(56235) - 2140)^(1/3))/(3 5^(2/3)) - 17/(3 (5 (2140 - 9 sqrt(56235)))^(1/3))

Bring each solution to a common denominator and simplify:

x = 0 or y = ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3))/(15 (2140 - 9 sqrt(56235))^(1/3)) or y = 1/15 (17 5^(2/3) ((-1)/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) or y = -(2140 - 9 sqrt(56235))^(1/3)/(3 5^(2/3)) - 17/(3 (5 (2140 - 9 sqrt(56235)))^(1/3))

Substitute back for x = y - 1/3:

x = 0 or x = 1/15 (2140 - 9 sqrt(56235))^(-1/3) ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3)) - 1/3 or x = 1/15 (17 5^(2/3) (-1/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) - 1/3 or x = -1/3 - 1/3 5^(-2/3) (2140 - 9 sqrt(56235))^(1/3) - 17/3 (5 (2140 - 9 sqrt(56235)))^(-1/3)

5 (2140 - 9 sqrt(56235)) = 10700 - 45 sqrt(56235):

x = 0 or x = ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3))/(15 (2140 - 9 sqrt(56235))^(1/3)) - 1/3 or x = 1/15 (17 5^(2/3) (-1/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) - 1/3 or x = -1/3 - (2140 - 9 sqrt(56235))^(1/3)/(3 5^(2/3)) - 17/(3 (10700 - 45 sqrt(56235))^(1/3))

6/x^2 + (2 x - 8)/(x + 5) ⇒ 6/0^2 + (2 0 - 8)/(5 + 0) = ∞^~

2/x - 3 ⇒ 2/0 - 3 = ∞^~:

So this solution is incorrect

6/x^2 + (2 x - 8)/(x + 5) ≈ -3.83766

2/x - 3 ≈ -3.83766:

So this solution is correct

6/x^2 + (2 x - 8)/(x + 5) ≈ -2.44783 + 1.13439 i

2/x - 3 ≈ -2.44783 + 1.13439 i:

So this solution is correct

6/x^2 + (2 x - 8)/(x + 5) ≈ -2.44783 - 1.13439 i

2/x - 3 ≈ -2.44783 - 1.13439 i:

So this solution is correct

The solutions are:

Answer: x = ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3))/(15 (2140 - 9 sqrt(56235))^(1/3)) - 1/3 or x = 1/15 (17 5^(2/3) (-1/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) - 1/3 or x = -1/3 - 17/(3 (10700 - 45 sqrt(56235))^(1/3)) - (2140 - 9 sqrt(56235))^(1/3)/(3 5^(2/3))

4 0

3 years ago

Spaceship Earth, a spherical attraction at Walt Disney worlds epcot center, has a diameter of 50 meters. Find the surface area o

gladu [14]

Answer:

The surface area of the structure is $SA=2,500\pi\ m^{2}$

Step-by-step explanation:

we know that

The surface area of a sphere is equal to

$SA=4\pi r^{2}$

In this problem we have

$r=50/2=25\ m$ -----The radius is half the diameter

substitute

$SA=4\pi (25)^{2}$

$SA=2,500\pi\ m^{2}$

5 0

3 years ago

What is the the y intercept in <br> y=-2+b

gayaneshka [121]

Your y-intercept is b bc you don’t have any thing yes but your Mx= -2

3 0

3 years ago

**PLEASE HELP I NEED IT WITHIN 20 MINUTES**

Genrish500 [490]

In mathematics, the trigonometric functions (also called circular functions, angle functions or goniometric functions[1][2]) are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others. They are among the simplest periodic functions, and as such are also widely used for studying periodic phenomena, through Fourier analysis.

Basis of trigonometry: if two right triangles have equal acute angles, they are similar, so their side lengths are proportional. Proportionality constants are written within the image: sin θ, cos θ, tan θ, where θ is the common measure of five acute angles.

The trigonometric functions most widely used in modern mathematics are the sine, the cosine, and the tangent. Their reciprocals are respectively the cosecant, the secant, and the cotangent, which are less used. Each of these six trigonometric functions has a corresponding inverse function (called inverse trigonometric function), and an equivalent in the hyperbolic functions as well.[3]

The oldest definitions of trigonometric functions, related to right-angle triangles, define them only for acute angles. To extending these definitions to functions whose domain is the whole projectively extended real line, geometrical definitions using the standard unit circle (i.e., a circle with radius 1 unit) is often used. Modern definitions express trigonometric functions as infinite series or as solutions of differential equations. This allows extending the domain of sine and cosine functions to the whole complex plane, and the domain of the other trigonometric functions to the complex plane (from which some isolated points are removed).

Contents

Right-angled triangle definitions Edit

A right triangle always includes a 90° (π/2 radians) angle, here labeled C. Angles A and B may vary. Trigonometric functions specify the relationships among side lengths and interior angles of a right triangle.

Plot of the six trigonometric functions, the unit circle, and a line for the angle θ = 0.7 radians. The points labelled 1, Sec(θ), Csc(θ) represent the length of the line segment from the origin to that point. Sin(θ), Tan(θ), and 1 are the heights to the line starting from the x-axis, while Cos(θ), 1, and Cot(θ) are lengths along the x-axis starting from the origin.

In this section, the same upper-case letter denotes a vertex of a triangle and the measure of the corresponding angle; the same lower case letter denotes an edge of the triangle and its length.

Given an acute angle A = θ of a right-angled triangle, the hypotenuse h is the side that connects the two acute angles. The side b adjacent to θ is the side of the triangle that connects θ to the right angle. The third side a is said to be opposite to θ.

If the angle θ is given, then all sides of the right-angled triangle are well-defined up to a scaling factor. This means that the ratio of any two side lengths depends only on θ. Thus these six ratios define six functions of θ, which are the trigonometric functions. More precisely, the six trigonometric functions are:[4][5]

sine

{\

4 0

3 years ago

What is the solution to the given inequality? -2x + 4 < 12